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Question Number 117482 by Don08q last updated on 12/Oct/20

If cosec^(−1) x = cot^(−1) y, show that (d^2 y/dx^2 ) + (1/y^3 ) = 0

$$\:\mathrm{If}\:\:\mathrm{cosec}^{−\mathrm{1}} {x}\:\:=\:\:\mathrm{cot}^{−\mathrm{1}} {y},\:\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:\:+\:\:\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\:\:=\:\:\mathrm{0} \\ $$

Answered by TANMAY PANACEA last updated on 12/Oct/20

y=cot(cosec^(−1) x) θ=cosec^(−1) x→cosecθ=x x^2 =1+cot^2 θ→cotθ=(√(x^2 −1)) y=(√(x^2 −1)) (dy/dx)=(1/(2(√(x^2 −1))))×2x=(x/( (√(x^2 −1)))) (d^2 y/dx^2 )=(((√(x^2 −1)) ×1−x×(1/(2(√(x^2 −1))))×2x)/(x^2 −1)) =((x^2 −1−x^2 )/((x^2 −1)^(3/2) ))=((−1)/y^3 ) proved

$${y}={cot}\left({cosec}^{−\mathrm{1}} {x}\right) \\ $$$$\theta={cosec}^{−\mathrm{1}} {x}\rightarrow{cosec}\theta={x} \\ $$$${x}^{\mathrm{2}} =\mathrm{1}+{cot}^{\mathrm{2}} \theta\rightarrow{cot}\theta=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$${y}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}×\mathrm{2}{x}=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:×\mathrm{1}−{x}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}×\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{{x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{−\mathrm{1}}{\boldsymbol{{y}}^{\mathrm{3}} }\:\boldsymbol{{proved}} \\ $$

Commented by $@y@m last updated on 12/Oct/20

Nice. I just want to add that there is no use of writing first line.

Commented by TANMAY PANACEA last updated on 13/Oct/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Don08q last updated on 13/Oct/20

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

Answered by $@y@m last updated on 13/Oct/20

Let cosec^(−1) x = cot^(−1) y=θ x=cosec θ ∣ y=cot θ (dx/dθ)=−cosec θcot θ ∣ (dy/dθ)=−cosec^2 θ (dy/dx)=((cosec θ)/(cot θ)) (dy/dx)= sec θ .....(1) (d^2 y/dx^2 ) =(d/dx)(sec θ)=(d/dθ)(sec θ)×(dθ/dx) (d^2 y/dx^2 ) =sec θtan θ×(1/(−cosec θcot θ)) (d^2 y/dx^2 ) =(1/(−cot^3 θ))=−(1/y^3 ) (d^2 y/dx^2 )+(1/y^3 )=0

$$\:{Let}\:\mathrm{cosec}^{−\mathrm{1}} {x}\:\:=\:\:\mathrm{cot}^{−\mathrm{1}} {y}=\theta \\ $$$${x}=\mathrm{cosec}\:\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{y}=\mathrm{cot}\:\theta \\ $$$$\frac{{dx}}{{d}\theta}=−\mathrm{cosec}\:\theta\mathrm{cot}\:\theta\:\:\:\:\mid\:\frac{{dy}}{{d}\theta}=−\mathrm{cosec}\:^{\mathrm{2}} \theta \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{cosec}\:\theta}{\mathrm{cot}\:\theta} \\ $$$$\frac{{dy}}{{dx}}=\:\mathrm{sec}\:\theta\:.....\left(\mathrm{1}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\frac{{d}}{{dx}}\left(\mathrm{sec}\:\theta\right)=\frac{{d}}{{d}\theta}\left(\mathrm{sec}\:\theta\right)×\frac{{d}\theta}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\mathrm{sec}\:\theta\mathrm{tan}\:\theta×\frac{\mathrm{1}}{−\mathrm{cosec}\:\theta\mathrm{cot}\:\theta} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{−\mathrm{cot}\:^{\mathrm{3}} \theta}=−\frac{\mathrm{1}}{{y}^{\mathrm{3}} } \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{\mathrm{1}}{{y}^{\mathrm{3}} }=\mathrm{0} \\ $$

Commented by Don08q last updated on 13/Oct/20

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

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